这篇博客详细解答了EE414课程的多项式插值问题，包括拉格朗日插值、牛顿插值等方法。通过对Q1到Q5的解答，讨论了不同模型的精度与计算复杂度，指出在需要高效计算时，牛顿模型优于拉格朗日模型，但在追求精确性时，拉格朗日模型更优。
Q1
First we can see the normal equations are:
(
∑
x
i
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c
+
(
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x
i
2
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m
=
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x
i
y
i
n
c
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m
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i
\left( \sum x_i \right) c + \left( \sum x^2_i \right) m = \sum x_i y_i \\ n c + \left( \sum x_i \right) m = \sum y_i \\
(∑xi)c+(∑xi2)m=∑xiyinc+(∑xi)m=∑yi
And the draw the rgresion table:
| i i i | x i x_i xi | y i y_i yi | x i 2 x^2_i xi2 | x i y i x_iy_i xiyi |
|---|---|---|---|---|
| 0 | -95 | -708 | 9025 | 67260 |
| 1 | -55 | -483 | 3025 | 26565 |
| 2 | -16 | -132 | 256 | 2112 |
| 3 | 24 | 254 | 576 | 6096 |
| ∑ \sum ∑ | -142 | -1069 | 12882 | 102033 |
So we can get:
{
−
142
c
+
12882
m
=
102033
4
c
−
142
m
=
−
1069
⇒
{
c
=
22.887
m
=
8.173
F
(
x
)
=
8.173
x
+
22.887
y
^
=
f
(
2
)
=
8.173
×
2
+
22.887
=
39.233
\left\{ \begin{aligned} -142 c + 12882 m &= 102033\\ 4 c - 142 m &= -1069 \\ \end{aligned} \right. \Rightarrow \left\{ \begin{aligned} c &= 22.887\\ m &= 8.173 \\ \end{aligned} \right. \\ F(x) = 8.173x + 22.887 \\ \hat{y} = f(2) = 8.173\times 2 + 22.887 = 39.233
{−142c+12882m4c−142m=102033=−1069⇒{cm=22.887=8.173F(x)=8.173x+22.887y^=f(2)=8.173×2+22.887=39.233
Q2
First we draw the divde difernce table:
| i i i | x i x_i xi | F ( x i ) F(x_i) F(xi) | 1 s t D D 1^{st}DD 1stDD | 2 n d D D 2^{nd}DD 2ndDD | 3 r d D D 3^{rd}DD 3rdDD |
|---|---|---|---|---|---|
| 0 | -95 | -708 | 5.625 | 0.04272 | − 2.8986 × 1 0 − 4 -2.8986\times 10^{-4} −2.8986×10−4 |
| 1 | -55 | -483 | 9 | 0.00823 | |
| 2 | -16 | -132 | 9.65 | ||
| 3 | 24 | 254 |
So we can get:
P
3
(
x
)
=
−
708
+
5.625
(
x
+
95
)
+
0.04272
(
x
+
95
)
(
x
+
55
)
−
2.8986
×
1
0
−
4
(
x
+
95
)
(
x
+
55
)
(
x
+
16
)
y
^
=
P
3
(
2
)
≈
44.9764
\begin{aligned} P_3(x) &= -708 + 5.625(x+95) + 0.04272(x+95)(x+55) -2.8986\times 10^{-4}(x+95)(x+55)(x+16) \\ \hat{y} &= P_3(2) \approx 44.9764 \\ \end{aligned}
P3(x)y^=−708+5.625(x+95)+0.04272(x+95)(x+55)−2.8986×10−4(x+95)(x+55)(x+16)=P3(2)≈44.9764
Q3
First we draw the datable:
| i i i | x i x_i xi | y i y_i yi |
|---|---|---|
| 0 | -95 | -708 |
| 1 | -55 | -483 |
| 2 | -16 | -132 |
| 3 | 24 | 254 |
Because
l
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∏
j
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0
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i
n
x
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j
x
i
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x
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l_i = \prod_{j=0,j\ne i}^n \frac{x-x_j}{x_i-x_j}
li=j=0,j=i∏nxi−xjx−xj, we can caluate:
l
0
=
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1
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x
2
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(
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x
3
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(
x
0
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(
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=
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55
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(
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16
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(
x
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24
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(
−
95
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55
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(
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95
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16
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(
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95
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24
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l
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=
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95
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16
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24
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55
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95
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(
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(
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24
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l
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95
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55
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x
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24
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95
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55
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l
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(
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0
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1
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3
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=
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95
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x
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55
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16
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(
24
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95
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(
24
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55
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(
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P
3
(
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)
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y
0
l
0
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y
1
l
1
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y
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l
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+
y
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l
3
=
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708
−
376040
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+
55
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x
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16
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x
−
24
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+
−
483
123240
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95
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16
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132
−
123240
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95
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254
376040
(
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95
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55
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(
x
+
16
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y
^
=
P
3
(
2
)
≈
44.9846
\begin{aligned} l_0 &= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)} = \frac{(x+55)(x+16)(x-24)}{(-95+55)(-95+16)(-95-24)} \\ l_1 &= \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} = \frac{(x+95)(x+16)(x-24)}{(-55+95)(-55+16)(-55-24)} \\ l_2 &= \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)} = \frac{(x+95)(x+55)(x-24)}{(-16+95)(-16+55)(-16-24)} \\ l_3 &= \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} = \frac{(x+95)(x+55)(x+16)}{(24+95)(24+55)(24+16)} \\ P_3(x) &= y_0 l_0 + y_1 l_1 + y_2 l_2 + y_3 l_3 \\ &= \frac{-708}{-376040}(x+55)(x+16)(x-24) + \frac{-483}{123240}(x+95)(x+16)(x-24) + \frac{-132}{-123240}(x+95)(x+55)(x-24) + \frac{254}{376040}(x+95)(x+55)(x+16) \\ \hat{y} &= P_3(2) \approx 44.9846 \\ \end{aligned}
l0l1l2l3P3(x)y^=(x0−x1)(x0−x2)(x0−x3)(x−x1)(x−x2)(x−x3)=(−95+55)(−95+16)(−95−24)(x+55)(x+16)(x−24)=(x1−x0)(x1−x2)(x1−x3)(x−x0)(x−x2)(x−x3)=(−55+95)(−55+16)(−55−24)(x+95)(x+16)(x−24)=(x2−x0)(x2−x1)(x2−x3)(x−x0)(x−x1)(x−x3)=(−16+95)(−16+55)(−16−24)(x+95)(x+55)(x−24)=(x3−x0)(x3−x1)(x3−x2)(x−x0)(x−x1)(x−x2)=(24+95)(24+55)(24+16)(x+95)(x+55)(x+16)=y0l0+y1l1+y2l2+y3l3=−376040−708(x+55)(x+16)(x−24)+123240−483(x+95)(x+16)(x−24)+−123240−132(x+95)(x+55)(x−24)+376040254(x+95)(x+55)(x+16)=P3(2)≈44.9846
Q4
\quad
First of al, in the three models, the y value caluated by the Lagrange model is very close to the true value, and when x equals 2, the estimated
y
^
\hat{y}
y^ value of the models is also very accurate. Howevr, the caluation amount of the Lagrange model is much larger than that of other two models.
\quad
Then, in the the Newton’s models, the y value caluated is very close to the true value in the former data, but it deviates from true value quickly in the later data. When x equals 2, the estimated
y
^
\hat{y}
y^ value of the models is very different with truth. Howevr, the caluation amount of the Newton’s model is much easier than Lagrange model.
\quad
So I think when you need estimated
y
^
\hat{y}
y^ value in smaller amount, the Newton’s model is better. But if you need a accurate model, the Lagrange mode is the best.
x = -100: 25; %%最低，步长，最高
y1 = 8.173*x + 22.887;
y2 = -708 + 5.625.*(x+95) + 0.04272.*(x+95).*(x+55) -2.8986.* 10^-4.*(x+95).*(x+55).*(x+16);
y3 = 708.*(x+55).*(x+16).*(x-24)./376040 - 483.*(x+95).*(x+16).*(x-24)./123240 + 132.*(x+95).*(x+55).*(x-24)./123240 + 254.*(x+95).*(x+55).*(x+16)./ 376040;
figure %%新建一个幕布
hold on;
plot(x, y1,'-');
plot(x, y2,'--');
plot(x, y3,'-.');
hold on;
scatter(-95, -708,'o','filled','b');
scatter(-55, -483,'o','filled','b');
scatter(-16, -132,'o','filled','b');
scatter(24, 254,'o','filled','b');
scatter(2, 39.233,'x','r');
scatter(2, 44.9764,'*','r');
scatter(2, 44.9846,'o','r');
title('Q4')%%加上标题
xlabel('x')%%x标签
ylabel('y')%%y标签
Q5
The data point is:
px = [3, 3.4, 3.8, 4.2, 4.6, 5, 5.4, 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2];
py = [3.5, 2.4, 0.55, -0.91, -0.41, -1.3, -1.1, -1.7, -0.16, 0.17, -0.03, 1.7, 4.4, 6.2];
figure %%新建一个幕布
hold on;
for i = 1:14
scatter(px(i), py(i),'o','filled','b');
end
title('Q5') %% 加上标题
xlabel('x') %% x标签
ylabel('y') %% y标签
\quad
Follow the observe of these items of data, I chose degree 13-order polynomial to regress these data.
\quad
So I use matlb to achiev this:
px = [3, 3.4, 3.8, 4.2, 4.6, 5, 5.4, 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2];
py = [3.5, 2.4, 0.55, -0.91, -0.41, -1.3, -1.1, -1.7, -0.16, 0.17, -0.03, 1.7, 4.4, 6.2];
len = 541;
x = linspace(2.9,8.3,len);
y = linspace(0,0,len);
for i = 1:14
z = linspace(1,1,len);
for j = 1:14
if j ~= i
for k = 1:len
z(k) = z(k) * (x(k)-px(j))/(px(i)-px(j));
end
else
continue;
end
end
for k = 1:len
y(k) = y(k) + py(i) * z(k);
end
end
figure %% 新建一个幕布
plot(x,y);
hold on;
for i = 1:14
scatter(px(i), py(i),'o','filled','b');
end
title('Q5') %% 加上标题
xlabel('x') %% x标签
ylabel('y') %% y标签
Conclusion:
\quad
Since there is 14 points, so when we choose the Lagrange module, the polynomial is 13-order. The results are found to be good through matlab simulation, reaching the expected results.
Q5 Method.2
px = [3, 3.4, 3.8, 4.2, 4.6, 5, 5.4, 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2];
py = [3.5, 2.4, 0.55, -0.91, -0.41, -1.3, -1.1, -1.7, -0.16, 0.17, -0.03, 1.7, 4.4, 6.2];
len = 541; %% 给定画图x的取值个数
x = linspace(2.9,8.3,len); %% 绘图x
r = length(px);
P = zeros(r,r); %% 制表
for i = 1:r
P(i,1) = py(i);
end
for i = 1: r-1 %% DD阶数
for j = 1: r-i
P(j,i+1) = (P(j+i,i)-P(j,i)) / (px(j+i)-px(j)); %% 得到i阶DD
end
end
e_y = ones(r,len); %% 建立计算的一组基
for i = 1: r-1
e_y(i+1,:) = e_y(i,:).*(x-px(i)); %% 下一组基等于前一组乘以x-p(i)
end
y = zeros(1,len); %% x对应的y值
for i = 1: r
y = y + P(1,i).*e_y(i,:); %% 线表计算结果
end
%% y = P(1,:)*e_y; %% x对应的y值，一步到位
figure %% 新建一个幕布
plot(x,y);
hold on;
for i = 1: r
scatter(px(i), py(i),'o','filled','b');
end
title('Q5') %% 加上标题
xlabel('x') %% x标签
ylabel('y') %% y标签
该代码Q5的数据拟合度很差，但该方法对应Q2的拟合度是很好的。我认为Method2在一定范围内是拟合度很高的，相较于拉格朗日插值法节省了相当多的计算过程。当数据点较多时，个人强烈推荐拉格朗日插值法来拟合数据。
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